I know that for every infinite dimensional Banach space $X$ admits a linear and discontinuous operator. To prove it, I took a sequence $\{e_n\}_{n\geq1}$ of linearly independent vectors of $X$, then I propose the operator $T:X\to \mathbb{R}$ defined as $T(e_n)=n||e_n||.$
Asumming Axiom of Choice, I can complete $\{e_n\}_{n\geq1}$ to a basis, and extend $T$ to a linear operator $\lambda$ such that, $\lambda(e_n)=T(e_n)$ and $\lambda(E_k)=0$, if $E_k$ is an element of the basis but different from every $e_n$. So I claim the operator $\lambda$ is a linear unbounded operator, i.e. a linear non-continuous operator.
My question is... How can I prove that the operator $\lambda$ indeed is a closed operator?
Every comment would be appreciated, even if I made a mistake writing my "proof" of the existence of a discontinuous linear operator.
I was trying to prove it but forgot elemental things about closed operators, blinded by the necessity of proving. Via definition: Isn't closed because, $e_n/\sqrt{n}$ is a convergent sequence but $\lambda(e_n/\sqrt{n}$ isn't. (ANSWER BY @Rishi Sonthalia, thanks).
Via closed graph theorem: Isn't closed because, if it is closed, then $\lambda$ must be bounded but we already prove that $\lambda$ indeed is unbounded. (ANSWER BY @KaviRamaMurthy).